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A cannonball is tired with an initial speed of 40m/s and a launch angle of 30degrees from a cliff that is 25m tall. a. What is the flight time of the cannonball b. How far away does the cannonball land c. Using two graphs, plot the horizontal speed of the projectile versus time and the vertical speed of the projectile versus time (from the initial launch of the projectile to the instant i strikes the ground).

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Answer:

Part a)

[tex]t = 5.1 s[/tex]

Part b)

[tex]x = 176.7 m[/tex]

Explanation:

Initial speed of the launch is given as

[tex]v_i = 40 m/s[/tex]

angle = 30 degree

Now the two components of the velocity

[tex]v_x = v_i cos30[/tex]

[tex]v_x = 34.6 m/s[/tex]

similarly we have

[tex]v_y = v_i sin30[/tex]

[tex]v_y = 20 m/s[/tex]

Part a)

Now to find the time of flight we will have

[tex]\Delta y = v_y t + \frac{1}{2}at^2[/tex]

[tex]-25 = 20 t -\frac{1}{2}(9.81)t^2[/tex]

[tex]t = 5.1 s[/tex]

Part b)

Distance at which the ball will land is given as

[tex]x = v_x t[/tex]

[tex]x = 34.6 (5.1)[/tex]

[tex]x = 176.7 m[/tex]

Part c)

Since x direction there is no acceleration so it will move with uniform speed while in y direction it will accelerate due to gravity

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